Micro4ProbSol9

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

0

Chapter 9

9.1 (a)

( )( )( )

2 1

3 31 10 10 50O d

d d

v A v v

A A− −

= −

= − − ⇒ =

(b)

( )32 2

2

1 500 10 1 0.5 500

3 mV

v v

v

−= − = + =

=

(c)

( )1 1

1

5 500 1 500 4950.990 V

v vv

= − ⇒ ==

(d) 0Ov =

(e)

( )( )2

2

2

3 500 0.5

250 3 5000.506 V

v

vv

− = − −

− − == −

______________________________________________________________________________________ 9.2

(a) 442 102

102 −×−=

−==

od

O

Aυ

υ V

Iυυ ⋅⎟⎠⎞

⎜⎝⎛+

=200011

2

4002.02001

1102 4 −=⇒⋅⎟⎠⎞

⎜⎝⎛=×− −

II υυ V

(b) Iυυ ⋅⎟⎠⎞

⎜⎝⎛+

=200011

2

( ) 32 109995.02

20011 −×=⎟

⎠⎞

⎜⎝⎛=υ V

( ) 5.1000109995.01 32 =⇒×=== −

odododO AAA υυ______________________________________________________________________________________ 9.3

(a) V ( ) ( )( ) 50010.20000.2105 312 −=−×=−= υυυ odO A

(b) ( )12 υυυ −= odO A

( )( ) 00265.30025.3102000.3 114 =⇒−×=− υυ V

(c) ( )12 υυυ −= odO A

( )( ) 43 1091001.001.080.1 ×=⇒×−−= −odod AA

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.4

( )

( )

25

0.790 0.8025

0.9875 2524.6875 0.0125

1975 K

iid I

i

i

i

i i

i

i

Rv v

R

RR

R RR

R

⎛ ⎞= ⎜ ⎟+⎝ ⎠

⎛ ⎞= ⎜ ⎟+⎝ ⎠

+ ==

=

______________________________________________________________________________________ 9.5

(a) 1020200

1

2 −=−

=−

=RR

Aυ

(b) 340120

−=−

=υA

(c) 14040

−=−

=υA

______________________________________________________________________________________ 9.6

200 1020

and for each case20 k

⎫= − = − ⎪⎪⎬⎪= Ω ⎪⎭

v

i

A

R

______________________________________________________________________________________ 9.7 a.

1

100 1010

10 k

= − = −

= = Ω

v

i

A

R R b.

510

100100−=−=υA

k 101 == RRi Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ c.

100 510 10

10 10 20 K

v

i

A

R

= − = −+

= + = ______________________________________________________________________________________ 9.8

(a) 200102040

2622

2 =⇒×

=⇒−

=−

RRR

i Oυ kΩ

67.1620012 111

2 =⇒−

=−⇒−

= RRR

RAυ kΩ

(b) μυ

5.7102005.100

2132

2 −==⇒×−

=−

= iiR

i O A

125.012

5.1−=⇒

−+

== IO

I Aυ

υυ

υ

V

______________________________________________________________________________________ 9.9

2

1v

RA

R= −

(a) 10vA = −

(b) 1vA = −

(c) 0.20vA = −

(d) 10vA = −

(e) 2vA = −

(f) 1vA = −

______________________________________________________________________________________ 9.10

(a) 2002003 211

2 =⇒−

=−

=− RRR

RkΩ , 67.661 =R kΩ

(b) 2002008 211

2 =⇒−

=−

=− RRR

RkΩ , 251 =R kΩ

(c) 20020020 211

2 =⇒−

=−

=− RRR

RkΩ , 101 =R kΩ

(d) 100200

5.0 22

1

2 =⇒−

=−

=− RR

RR

kΩ , 2001 =R kΩ

______________________________________________________________________________________ 9.11

(a) 5105025.0

1611

1 =⇒×

=⇒=−

RRR

i Iυ kΩ

5.325

5.6 22

1

2 =⇒−

=−⇒−

= RR

RR

Aυ kΩ


(b) 6154.05.6

4=

−−

==υ

υυ

AO

I V

123.05.32

421 === ii mA

______________________________________________________________________________________ 9.12

(a) 1

2

RR

A−

=υ

2525

20 12 =⇒

−=− R

Rk , Ω 5002 =R kΩ

(b) 50100020 11

=⇒−

=− RR

kΩ , 12 =R MΩ

(c) For (a), μυ

825

2.01

11 −=⇒

−== i

Ri I A

For (b), μυ

450

2.01

11 −=⇒

−== i

Ri I A

______________________________________________________________________________________ 9.13 a.

2 2 2

1 1 1

2 2

1 1

1.05 1.1050.95

0.95 0.9051.05Deviation in gain is +10.5% and 9.5%

⎛ ⎞= ⇒ = ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

−

vR R RAR R R

R RR R

b.

2 2 2

1 1 1

1.01 0.991.02 0.980.99 1.01

Deviation in gain 2%

vR R RAR R R

⎛ ⎞ ⎛ ⎞⇒ = ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= ±

2

1

RR

______________________________________________________________________________________ 9.14

(a) (i) ( ) 320.0115

1

2 =−−

=⋅−

= IO RR

υυ V

(ii) 20.015

30

22 −=

−=

−=

Ri Oυ mA

75.043===

L

OL R

iυ

mA

( ) 95.020.075.02 =−−=⇒=+ OLO iiii mA


(b) (i) ( ) 75.005.0115

−=−

=Oυ V

(ii) ( ) 05.015

75.002 =

−−=i mA

1875.0475.0

−=−

=Li mA

2375.005.01875.02 −=−−=−= iii LO mA

(c) (i) ( tO ωυ sin8115−

= ) (mV) ⇒ tO ωυ sin12.0−= (V)

(ii) tit

i ωω

sin815sin12.0

22 =⇒= ( μ A)

tit

i LL ωω

sin304sin12.0

−=⇒−

= ( μ A)

tiii LO ωsin382 −=−= ( μ A) ______________________________________________________________________________________ 9.15

2

1 5

30 2.5% 29.25 30.75

v

v v

RA

R RA A

= −+

= − ± ⇒ ≤ ≤

2 2

1 1

So 29.25 and 30.752 1

R RR R

= =+ +

( ) (1 1We have 29.25 2 30.75 1R R+ = + )

1 2Which yields 18.5 and 599.6 R k R= Ω = kΩ

For 25 , then 0.731 0.769 I Ov mV v= ≤ ≤ V

______________________________________________________________________________________ 9.16

( ) 2.115.01080

1

21 =−

−=⋅

−= IO R

Rυυ V

( ) 62.120100

13

4 −=−

=⋅−

= OO RR

υυ V

( ) 1510

15.021

121 −==⇒

−=== ii

Rii Iυ μ A

6020

2.143

3

143 ==⇒=== ii

Rii Oυ μ A

At 1Oυ : ( ) 7515601312 =−−=⇒=+ OO iiii μ A; Out of Op-Amp At Oυ : 6042 == iiO μ A; Into Op-Amp ______________________________________________________________________________________ 9.17

1003

4

1

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RR

RR

I

O

υυ

For 50=Iυ mV, ( )( ) 505.0100 ==Oυ V


If μ504 =i A, 10010505

464 =⇒×

=−

RR kΩ

Set k 101 =R Ω

Then 1010010

1003

2

3

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

RR

RR

I

O

υυ

Set k , k 1002 =R Ω 103 =R Ω______________________________________________________________________________________ 9.18

3005

6

3

4

1

2 −=⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RR

RR

RR

Aυ

For 6=Oυ V, set μυ

606

6 ⇒=R

i O A 10066

6

=⇒= RR

kΩ

Set kΩ so that 2006 =R 306 =i μ A Set kΩ 201 =R

Now ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

53

42 20020

300RR

RR

For example, set kΩ and 1002 =R 205 =R kΩ

Then ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

3

4

3

4 5020

20020

100300RR

RR

Or 63

4 =RR

, set k and 203 =R Ω 1204 =R kΩ

______________________________________________________________________________________ 9.19

(a) ( ) 8.840.0122

1

2 =−⎟⎠⎞

⎜⎝⎛−=⋅

−= IO R

Rυυ V

(b) ( )( )

8993.2123

10511

122

111

1

31

21

2 −=

⎥⎦⎤

⎢⎣⎡

×+

⋅−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

⋅−

=

RR

A

RR

A

od

υ

( )( ) 7597.840.08993.21 =−−=Oυ V (c) ( )( ) 956.2122998.0 −=−=υA

( )( )

4101477.12311

122956.21 ×=⇒

⎥⎦

⎤⎢⎣

⎡+

⋅−=− od

od

A

A

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.20 (a)

( )

2

1 2

1

3

111 1

100 1125 1 5

5 103.9960

v

od

v

RA

R RA R

A

= − ⋅⎡ ⎤⎛ ⎞+ +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= − ⋅⎡ ⎤+⎢ ⎥×⎣ ⎦

= −

(b) ( )3.9960 1.00 3.9960 VO Ov v= − ⇒ = −

(c) 4 3.9960 100% 0.10%

4−

× =

(d)

( )( )

2 1 1

1 3

1

3.99605 10

0.7992 mV

O od od

O

od

v A v v A v

vv

Av

+

= − = −

− −= − =

×=

______________________________________________________________________________________ 9.21

(a) 98431.9

101001

10711

110100

3

−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

×+

⋅−

=υA

7011.09843.97

−=−

==υ

υυ

AO

I V

11077

131 −=⇒×−

=−

= υυ

υυAO mV

(b) ( ) 43

1

105.2102.05

×=×−−

=−

=−υ

υOodA

( )( )

9956.911

105.211

110

4

−=

⎥⎦⎤

⎢⎣⎡

×+

⋅−=υA

50022.09956.95

=−

−==

υ

υυ

AO

I V

______________________________________________________________________________________ 9.22

(a) 605050

5501

10501

2

3

4

3

1

2 −=⎟⎠⎞

⎜⎝⎛ ++

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛++

−=

RR

RR

RR

Aυ

(b) (i) 78.250505015100 4

4

=⇒⎟⎟⎠

⎞⎜⎜⎝

⎛++−=− R

RkΩ

(ii) 79.150505015150 4

4

=⇒⎟⎟⎠

⎞⎜⎜⎝

⎛++−=− R

RkΩ

_____________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.23 a.

3 32

1 4 2

1

3 32

4 2

2 3

44 4

1

500 k

80 1500

Set 500 k

500 50080 1 1 1 2 6.41 k

vR RRA

R R RR

R RRR R

R R

RR R

⎛ ⎞= − + +⎜ ⎟

⎝ ⎠= Ω

⎛ ⎞= + +⎜ ⎟

⎝ ⎠= = Ω

⎛ ⎞= + + = + ⇒ = Ω⎜ ⎟

⎝ ⎠ b.

( )( )1 2 1 2

6 32 2

4 44

3 2 4 3

For 0.05 V0.05 0.1 A

500 k0.1 10 500 10 0.05

0.05 7.80 A6.410.1 7.80 7.90 A

I

X

X

v

i i i i

v i R

vi iR

i i i i

μ

μ

μ

−

= −−

= = ⇒ = = −Ω

= − = − − × × =

= − = − ⇒ = −

= + = − − ⇒ = −

______________________________________________________________________________________ 9.24 (a)

2

1 1

1

5001000

0.5 K

vRAR R

R

− −= − = =

=

(b)

3 32

1 4 2

1 1

1

1

250 500 500 12501000 1250 250

1.25 K

vR RRA

R R R

R RR

⎛ ⎞−= + +⎜ ⎟

⎝ ⎠− −⎛ ⎞− = + + =⎜ ⎟

⎝ ⎠=

______________________________________________________________________________________ 9.25


1 2

2

3

I

IA I

A I

vi iR

vv i R R vR

v viR R

= =

⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

= − =

( )

( )

4 2 3

4

5

6 4 5

2 2

2 3

3 3

2 3 5

A A A I

IB A I

IB I

I I I

v v v vi i iR R R R

vv v i R v R vR

vv viR R R

v v vi i iR R R

= + = − − = − =

⎛ ⎞= − = − − = −⎜ ⎟⎝ ⎠

−= − = − =

= + = + =

I

00 6

53 8

From Figure 9.12 3

IB I

I

v

vvv v i R v RR v

A

⎛ ⎞= − = − − ⇒ = −⎜ ⎟⎝ ⎠

⇒ = −

______________________________________________________________________________________ 9.26

(a) 9978.9

202001

10511

120200

111

1

41

21

2 −=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

×+

⋅−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

⋅−

=

RR

A

RR

A

od

υ

(b) 4801056.09978.9

80.4=

−−

==υ

υυ

AO

I V

(c) ( ) μυυ

υ 96105

80.4141 =⇒

×−−

=−

=od

O

AV

(d) %022.0%1008.4

8.4801056.4=×

−

______________________________________________________________________________________ 9.27

(a) ( )( )

9992.02

105.211

11

111

1

31

21

2 −=

⎥⎦⎤

⎢⎣⎡

×+

⋅−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

⋅−

=

RR

A

RR

A

od

υ

(i) ( )( ) 7993605.08.09992.0 =−−=Oυ V

(ii) %08.0%1008.0

7993605.08.0≅×

−

(b) ( )( )

990099.02

10211

11

2

−=

⎥⎦⎤

⎢⎣⎡

×+

⋅−=υA

(i) ( )( ) 79208.08.0990099.0 =−−=Oυ V

(ii) %99.0%1009.079208.08.0

=×−

______________________________________________________________________________________


(a)

22

1 2

l O Oi

l

v v v Ri iR R v

= = = − ⇒ = −1R

(b)

22 1 3 3

1 1

23

1

1

Then 1

l Ol

L L

l

L

v v Ri i i i vR R R R

v RiR R

⎛ ⎞= = = + = + − ⋅⎜ ⎟

⎝ ⎠⎛ ⎞

= +⎜ ⎟⎝ ⎠

______________________________________________________________________________________ 9.29

( )

( )

3 1.max .max

3 1 4

2.max

1

2 2

1 1

2

0.1 110 0.09008

0.1 1 10

10 0.09008 111

So 111

+⎛ ⎞ ⎛ ⎞= ⋅ = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

= ⋅

= ⇒ =

= Ω

X X

O X

R RV V V

R R R

Rv VR

R RR R

R k

V

______________________________________________________________________________________ 9.30

33

22

11

IF

IF

IF

O RR

RR

RR

υυυυ ⋅−⋅−⋅−= 321 60120

20120

40120

III υυυ ⋅−⋅−⋅−=

321 263 IIIO υυυυ −−−= (a) ( ) ( ) ( ) 85.25.1210.0625.03 −=−−−−=Oυ V (b) ( ) ( ) ( ) 133.02.1225.0635.0 11 =⇒−−−−= II υυ V

______________________________________________________________________________________ 9.31

(a) ( ) 321321 625.025.6325.05.22.15.2 IIIIIIO υυυυυυυ −−−=++−=

Then 31

=RRF , 25.6

2

=RRF , 625.0

3

=RRF

3R is the largest resistor, so set 4003 =R kΩ Then k , k250=FR Ω 3.831 =R Ω , 402 =R kΩ

(b) ( ) ( ) ( ) 1875.02625.025.025.613 =−−−−=Oυ V

μυ

75.02501875.0

=⇒== FF

OF i

Ri A

______________________________________________________________________________________ 9.32 ( ) 2121 6232 IIIIO υυυυυ −−=+−=

Then 21

=RRF , 6

2

=RRF

For 11 −=Iυ V, 5.02 −=Iυ V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then ( ) ( ) 55.0612 =−−−−=Oυ V

Set μ80=Fi A 5.625=⇒== F

FF

O RRR

υkΩ

Then k , k25.311 =R Ω 42.102 =R Ω ______________________________________________________________________________________ 9.33

( ) ( ) ( )

( ) ( )( ) ( )( )( ) ( )

1

2 23

1 2 1 21 2

0.05 2 sin 2 0.0707sin 21 11 1 10

1001010 101 5

10 0.0707sin 2 2 1

0.707sin 2 2

I

I

F FO I I I I

O

O

v ft ft

f kHz T ms v T ms

R Rv v v v vR R

v ft V

v ft V

π π

π

π

= =

= ⇒ = ⇒ ⇒ = ⇒

= − ⋅ − ⋅ = − ⋅ − ⋅

= − − ±

= − − ±

______________________________________________________________________________________ 9.34

22

11

IF

IF

O RR

RR

υυυ ⋅−⋅−=

( ) ( )006.0100sin125.0004.0100sin5.021

−−+−=−R

tR

t ωω

Set ( ) 25sin125.0100sin5.0 11

=⇒−=− RtR

t ωω kΩ

We have ( ) ( )006.0100004.0100021

−−−=RR

5.376.025

4.00 22

=⇒+−= RR

kΩ

______________________________________________________________________________________ 9.35

(a) 321321 24

212

42 IIIII

IO υυυυυ

υυ −−−=⎥

⎦

⎤⎢⎣

⎡++−=

Then 21

1

=RRF , 4

2

=RRF , 2

3

=RRF

Set k2501 =R Ω , Then k125=FR Ω , 25.312 =R kΩ , 5.623 =R kΩ


(b) For 21 −=Iυ V, 02 =Iυ , 13 −=Iυ V

( ) ( ) ( ) 31204221

=−−−−−=Oυ V

For 21 =Iυ V, 5.02 =Iυ V, 03 =Iυ

( ) ( ) ( ) 3025.04221

−=−−−=Oυ V

Then 33 +≤≤− Oυ V

μυ

24125

3max

maxmax

=⇒== FF

OF i

Ri A

______________________________________________________________________________________ 9.36

33

22

11

IF

IF

IF

O RR

RR

RR

υυυυ ⋅−⋅−⋅−=

( ) ( ) ( )4sin5.0sin22sin6321

−−−+−=−RR

tRR

tRR

t FFF ωωω

We have ( ) ( ) 1331

2042 RRRR

RR FF =⇒=+−

Also ( ) ( )5.02621 R

RRR FF −−=−

For 6=Oυ V, μ120max

=Fi A 506=⇒= F

F

RR

kΩ

For 4max1 =Iυ V, μ120

max1 =i A 33.3341

1

=⇒= RR

kΩ and 66.662 13 == RR k Ω

Now ( )( ) ( )( ) 33.82533.33

100251005.0502506 222121

=⇒+=+=+= RRRRRR

kΩ

______________________________________________________________________________________ 9.37 a.

( ) ( ) ( ) ( )

( )

0 3 2 1 03 2 1 0

3 02 10

5 5 5

So 510 2 4 8 16

F F F F

F

R R R Rv a a a aR R R R

a aR a av

= − ⋅ − − ⋅ − − ⋅ − − ⋅ −

⎡ ⎤= + + +⎢ ⎥⎣ ⎦

5

b. 0

12.5 5 10 k10 2

FF

Rv R= = ⋅ ⋅ ⇒ = Ω

c.

i. 0 0

10 1 5 0.3125 V10 16

v v= ⋅ ⋅ ⇒ =

ii. ( )0 0

10 1 1 1 1 5 4.68710 2 4 8 16

v v⎡ ⎤= + + + ⇒ =⎢ ⎥⎣ ⎦5 V

______________________________________________________________________________________


(a) 2121 20200120

120

110

IIIIO υυυυυ −=⋅⎟⎠⎞

⎜⎝⎛−⋅⎟

⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ −=

(b) ( )( ) ( )( ) ttO ωωυ sin10005001000sin5025205200 ++=−−−= (mV) tO ωυ sin0.15.1 += (V) (c) For the 20 k resistor: Ω

125.020

5.2maxmax

=⇒= ii mA

For the 10 kΩ resistor:

( ) 5051

101 =⎟

⎠⎞

⎜⎝⎛=Oυ mV, μ5

1050

maxmax=⇒

Ω= i

kmV

i A

______________________________________________________________________________________ 9.39 For one-input

od

O

Aυ

υ −=1

F

OI

RRRRυυυυυ −

+=− 1

32

1

1

11

F

O

F

I

RRRRRRυ

υυ

−⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

111

3211

1

1

F

O

Fod

O

RRRRRAυυ

−⎥⎥⎦

⎤

⎢⎢⎣

⎡++−=

111

321

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++−=

321

11111RRRARRA odFFod

Oυ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⋅++−=

321

111RRR

RAAR

F

ododF

Oυ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

⋅⋅−=

P

F

od

IF

O

RR

A

RR

111

11

1

υυ where 321 RRRRP =

Therefore, for three-inputs ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅+⋅+⋅⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−= 3

32

21

1111

1I

FI

FI

F

P

F

od

O RR

RR

RR

RR

A

υυυυ

______________________________________________________________________________________


(a) 1115

150111

2 =⎟⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

RR

Aυ

(b) 450

1501 =⎟⎠⎞

⎜⎝⎛ +=υA

(c) 4.150201 =⎟

⎠⎞

⎜⎝⎛ +=υA

(d) 220201 =⎟

⎠⎞

⎜⎝⎛ +=υA

______________________________________________________________________________________ 9.41

(a) 141151

2

1

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+==

RR

RR

Aυ

For 5.7−=Oυ V 5.01 −=⇒υ V

μ120=i A 33.585.05.72

2

=⇒−

= RR

kΩ , 17.41 =R kΩ

(b) ( )( ) 75.325.015 ==Oυ V

μ6017.425.0

2121 ==⇒== iiii A

______________________________________________________________________________________ 9.42

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

1

21RR

Aυ

2131

2

1

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

, Set 2902 =R kΩ , 1451 =R kΩ

(b) 8191

2

1

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

, Set 2902 =R kΩ , 25.361 =R kΩ

(c) 291301

2

1

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

, Set 2902 =R kΩ , 101 =R kΩ

(d) 0111

2

1

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

, Set 02 =R , 2901 =R kΩ

______________________________________________________________________________________ 9.43

0

1 11 500 501B I odv v v A⎛ ⎞ ⎛= =⎜ ⎟ ⎜+⎝ ⎠ ⎝

iv⎞⎟⎠


a. ( )12.5 5 250.5

501od odA A⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

b. ( )0 0

15000 5 49.9 V501

v v⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

______________________________________________________________________________________ 9.44

0 2

0 1 2

50 20 40150 20 40 20 40

1.33 0.667

I I

I I

v v

v v v

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦= +

1v

______________________________________________________________________________________ 9.45 (a)

1 2 2 2 2

2 2

1 2 2 2

1 2 2

1 2

20 40 101001 350

2Now 2 2 4

2 7 73

6 3So 7 7

I I

O

I I

oI I

O I I

v v v v v

v v v

v v v v vv

v v v

v v v

− −+ =

⎛ ⎞= + =⎜ ⎟⎝ ⎠

− + − =

⎛ ⎞+ = = ⎜ ⎟⎝ ⎠

= ⋅ + ⋅

(b) ( ) ( )6 30.2 0.3 0.3 V

7 7O Ov v⎛ ⎞= + ⇒ =⎜ ⎟⎝ ⎠

(c) ( ) ( )6 30.25 0.4 42.86 mV

7 7O Ov v⎛ ⎞ ⎛ ⎞= + − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

______________________________________________________________________________________ 9.46

(a) IRRR

υυ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+

=43

42

IO

RRR

RRR

υυυ ⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+=⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

4

31

22

1

2

1

111

(b) 891

50251

1161

2

1

2

1

2 =⇒=⎟⎟⎠

⎞⎜⎜⎝

⎛+⇒

⎟⎠⎞

⎜⎝⎛ +⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

RR

Set k2002 =R Ω , k 251 =R Ω______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.47 (a)

( )501

1 50

111 1

11

O

I

O

I

Ov

I

v xv x

v x x xv x

vAv x

⎛ ⎞= +⎜ ⎟⎜ ⎟−⎝ ⎠

− +⎛ ⎞= + =⎜ ⎟− −⎝ ⎠

= =−

x

(b) 1 vA≤ ≤ ∞

(c) If x = 1, gain goes to infinity. ______________________________________________________________________________________ 9.48

(a) ( ) III

X RR

υυυ

υ 32 =+⎟⎟⎠

⎞⎜⎜⎝

⎛=

022

=−

++−

RRROXXIX υυυυυ

RRRRROI

X 22211

21 υυ

υ =−⎟⎠⎞

⎜⎝⎛ ++

RRROI

I 2223

υυυ =−⎟

⎠⎞

⎜⎝⎛

so 11=I

O

υυ

(b) For 25.0=Iυ V, 75.2=⇒ Oυ V (c) k , 30=R Ω 15.0−=Iυ V

For : 1R μ53015.0

⇒=i A

For : 2R μ5=i A 45.03 −== IX υυ V

For : 4R μ153045.0

⇒=i A

( )( ) 65.115.011 −=−=Oυ V

For : 3R μ2060

45.065.1⇒

−=i A

______________________________________________________________________________________


(a) 1O

I

vv

=

(b) From Exercise TYU9.7

2

1

2

1

2 1

5

1

11 1

But 0, 1 1 0.999993

1 11 11.5 10

O

I

od

O O

I I

od

RRv

v RA R

R Rv vv v

A

⎛ ⎞+⎜ ⎟

⎝ ⎠=⎡ ⎤⎛ ⎞+ +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦= = ∞

= = ⇒ =+ +

×

(b)

1Want 0.990 9911

Ood

I

od

vA

vA

= = ⇒ =+

______________________________________________________________________________________ 9.50

(a) ( ) ( OIododO AA )υυυυυ −=−= 12 ( ) IododO AA υυ =+1

9524.0

2011

111

1=

+=

+==

od

I

O

A

Aυυ

υ

(b) 995.0

20011

1=

+=υA

(c) 9995.0

200011

1=

+=υA

(d) 99995.0

2000011

1=

+=υA

______________________________________________________________________________________ 9.51

(a) ⎟⎟⎠

⎞⎜⎜⎝

⎛+==

1

211 1

RR

AI

O

υυ

υ

⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

1

222 1

RR

AI

O

υυ

υ

(b) ( ) 25.0206011 −=−⎟

⎠⎞

⎜⎝⎛ +=Oυ V

( ) 25.0206012 =−⎟

⎠⎞

⎜⎝⎛ +−=Oυ V


(c) ( ) 4.68.020601212

1

221 =⎟

⎠⎞

⎜⎝⎛ +=⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+=− IOO R

Rυυυ V

______________________________________________________________________________________ 9.52

(a) 1

= IL

viR

(b)

( ) ( )( )

( ) ( )( ) ( )

1 1

1

max 10 V 1 9 10So max 1 mA

Then max 1 9 max 9 V

O L L I L L L

OI L L

L

I L I

v i R v i R i Rv i i

i

v i R v

= + = +≅ = + =≅

≅ = ⇒ ≅

______________________________________________________________________________________ 9.53

(a) ( ) IIO υυυ 3333.04020

20=⋅⎟

⎠⎞

⎜⎝⎛

+=

(i) 1=Oυ V (ii) 67.1−=Oυ V

(b) ( ) IIO υυυ 3333.04020

20=⋅⎟

⎠⎞

⎜⎝⎛

+=

(i) 1=Oυ V (ii) 67.1−=Oυ V

(c) ( ) IIO υυυ 2222.0486

610101 =⋅⎟

⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛ +=

(i) 667.0=Oυ V (ii) 111.1−=Oυ V

______________________________________________________________________________________ 9.54 a.

( ) ( )

1 011 0

1

1 1 11

1

1

and and

1So

Then 1

in odF

od od

F F

Fin

od

v vvR i vi R

v A v v Ai

R Rv RRi A

−= = =

− − += =

= =+

1A v−

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

1 0

0

0

and 1

So 1

10 0.0099901 1001

10001001 0.009990

1000Want 0.9901001 0.009990

which y

S FS od

S in od

od SF S

od S in

Fin

od

SF S

S

S

S

R Ri i v AR R A

A Rv R iA R R

RRA

Rv R iR

RR

⎛ ⎞= = − ⋅⎜ ⎟+ +⎝ ⎠

⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

= = =+

⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ≤⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠

1i⋅

ields 1.099 kSR ≥ Ω

______________________________________________________________________________________ 9.55

( ), 0 8 mA

For max 8 V, Then 1 O C F C

O F

v i R iv R= ≤ ≤

= = kΩ

______________________________________________________________________________________ 9.56

10 so 1 10 kIv

i RR R

= = ⇒ = Ω

In the ideal op-amp, R1 has no influence.

Output voltage: 2

0 1 IRv vR

⎛ ⎞= +⎜ ⎟⎝ ⎠

v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. ______________________________________________________________________________________ 9.57

(a) 2R

i IL

υ−= ; ( ) 1

55

2 =−−

=R kΩ

Set 231

1RRR

RF = ; For example, set 101 == FRR kΩ , 13 =R kΩ

(b) ( )( ) 12.05 ==Lυ V 1υ=

6.010

15

1

121 −=

−−=

−==

Rii I υυ

mA


( )( ) 7106.0121 =−−=−= FO Riυυ V

61

17

33 =

−=

−=

Ri LO υυ

mA

111

24 ===

Ri Lυ mA

For the op-amp: ( ) 6.66.062332 =−−=−=⇒=+ iiiiii OO mA ______________________________________________________________________________________ 9.58 (a)

1 2 2 22

1 12

12

and ,

Then

Or 1

= = + = −

⎛ ⎞= − +⎜ ⎟

⎝ ⎠⎛ ⎞

= +⎜ ⎟⎝ ⎠

xD x F

FD

FD

vi i i i v i RR

Ri i iR

Ri iR

(b)

( )

1 11

2 2

2

5 5 1

12 1 1 11

For example, 5 , 55

= = ⇒ = Ω

⎛ ⎞= + ⇒ =⎜ ⎟

⎝ ⎠= Ω = Ω

I

F F

F

vR R ki

R RR R

R k R k ______________________________________________________________________________________ 9.59

(1) 2 3

X OXX

V vVIR R

−= +

(2) 1

0X OX

F

V vVR R

−+ =

1

2 3 3 1

0 2 3 3 1 3 2 1 3

1 3 2

1 2 3

1 2 3

1 3 2

2 1 31 3 2

From (2) 1

1 1 1Then (1) 1

1 1 1 1 1

or

1Note: If then , which corresp

FO X

FX X X

X F F

X

F

oF

FF o

Rv VR

RI V VR R R R

I R RV R R R R R R R R R

R R R RR R R

R R RRR R R R

R R R R R RR R R

⎛ ⎞= +⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞

= + − ⋅ +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= = + − − = −

−=

=−

= ⇒ = = ∞ onds to an ideal current source.

______________________________________________________________________________________


(a) kΩ ; 3031 =+= RRRid 1531 == RR kΩ

22515 423

4

1

2 ==⇒== RRRR

RR

kΩ

(b) ( )( ) 5.21025.0 === LLO Riυ V

1667.015

5.212 ===−

d

OII A

υυυ V

(c) ( ) ( ) 5.45.12.11512 −=−=−= IIdO A υυυ V

45.010

5.4−=

−==

L

OL R

iυ

mA

(d) ( )( ) 5105.0 ==Oυ V

333.0155

12 ===−d

OII A

υυυ V

667.1333.021 =−=Iυ V ______________________________________________________________________________________ 9.61

(a) 403

4

1

2 ==RR

RR

; Set k25042 == RR Ω , 25.631 == RR kΩ

(b) 253

4

1

2 ==RR

RR

; Set k25042 == RR Ω , 1031 == RR kΩ

(c) 53

4

1

2 ==RR

RR

; Set k25042 == RR Ω , 5031 == RR kΩ

(d) 5.03

4

1

2 ==RR

RR

; Set k12542 == RR Ω , 25031 == RR kΩ

______________________________________________________________________________________ 9.62

( ) ( )( ) ( )

4 32 2 22 1 2 1

1 4 3 1 1 3 4 1

2 1

3 4

2

We have

/ 11 or 11 / 1 /

Set 50 1 , 50 150 1 , 50 1

1 1111 11

O I I O I

O I

R RR R R Rv v v v vR R R R R R R R

R x R xR x R x

xv vxxx

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= + = −= − = +

⎡ ⎤⎢ ⎥⎡ + ⎤⎛ ⎞ ⎢ ⎥= + ⎜ ⎟⎢ ⎥ −− ⎛ ⎞⎢ ⎥⎝ ⎠⎣ ⎦ + ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

2Iv

( )( )

1

2 1

2 1

11

1 1 1 11 1 1 1

1 11 1

I

O I I

I I

x vx

x x x xv v vx x x x

x xv vx x

+⎛ ⎞− ⎜ ⎟−⎝ ⎠

⎡ ⎤− + +⎡ ⎤ + +⎛ ⎞= ⋅ −⎢ ⎥⎢ ⎥ ⎜ ⎟− + + − −⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦+ +⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠


( ) ( )( ) ( )

( )

1 2

2 1

3 4

2 1

2 1

1 2

For 0Set 50 1 50 1

50 1 50 1

1 1 1111 111

11

1 11 211 1 1

I I O

O I

I I

I I cm

O

cm

v v vR x R xR x R x

x xv vxx xx

xv vx

v v vx xv x x

v x x x

= ⇒ =

= + = −= + = −

⎛ ⎞⎜ ⎟+ +⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟+− −⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟

−⎝ ⎠+⎛ ⎞= − ⎜ ⎟−⎝ ⎠

= =

− − ++ −= − = =

− − −

Iv

( ) ( )( ) ( )

( )

2 1

3 4

2 1

Set 50 1 50 150 1 50 1

1 1 1111 111

111

1 1 21 1

O I

cm

cm

R x R xR x R x

x xv vxx xx

x vx

x x xAx x

= − = += − = +

⎛ ⎞⎜ ⎟− −⎛ ⎞ ⎛ ⎞= + −⎜ ⎟⎜ ⎟ ⎜ ⎟−+ +⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟+⎝ ⎠

−⎛ ⎞= −⎜ ⎟+⎝ ⎠+ − −

= =+ +

Iv

Worst common-mode gain2

1cmxAx

−=

− (b)

( )

( )

( )

( )

2 1

2 0.012For 0.01, 0.02021 1 0.01

2 0.02For 0.02, 0.04082

1 0.022 0.05

For 0.05, 0.10531 0.05

1 1For this condition, set , 1 V2 2

1 11 1 112 1 2 1

cm

cm

cm

I I d

d

xx Ax

x A

x A

v v v

x xxAx

−−= = = = −

− −−

= = = −−

−= = = −

−

= + = − ⇒ =

− + +⎡ + ⎤⎛ ⎞= + =⎜ ⎟⎢ ⎥− −⎝ ⎠⎣ ⎦

10

10

10

1 2 12 1 1

1.010For 0.01 1.010 20log 33.98 dB0.0202

1 1.0For 0.02, 1.020 20log 27.96 dB0.98 0.040821 1.0526For 0.05 1.0526 20log 20 dB

0.95 0.1053

d dB

d dB

d dB

x x x

x A C M R R

x A C M R R

x A C M R R

⎡ ⎤= ⋅ =⎢ ⎥ − −⎣ ⎦

= = = =

= = = = =

= = = = ≅

20

______________________________________________________________________________________


(a) ( ) =−= 1210 υυυO ( ) 48.14.110 −=− V

( )( ) 0127.010114.1

102

43 ==+

==RR

iiυ

mA

( ) 273.14.11110

1110

2 =⎟⎠⎞

⎜⎝⎛=⋅⎟

⎠⎞

⎜⎝⎛== υυυ YX V

0527.010

273.18.1121 =

−=

−==

Rii Xυυ

mA

(b) ( ) ( ) 42.36.31010 12 =−=−= υυυO V

( ) 273.36.31110

=⎟⎠⎞

⎜⎝⎛== YX υυ V

( )( ) 0327.0110

6.310112

43 ====υ

ii mA

00727.010

273.32.321 −=

−== ii mA

(c) ( )( ) 5.120.135.110 −=−−−=Oυ V

( )( ) 0123.0101135.1

43 −=−

== ii mA

( ) 227.135.11110

−=−⎟⎠⎞

⎜⎝⎛== YX υυ V

( ) 00273.010

227.12.121 =

−−−== ii mA

______________________________________________________________________________________ 9.64

(a) ( ) ( )( ) 2.912.1761 ==⋅+= BE II β mA

Ω=⇒= 6.1092.91

10 RR

(b) mA ( )( ) 2.202.0101 ==EI

495.02.20

10==R kΩ

(c) 74.541096.06

==EI mA

72.076

74.54==OI mA

(d) 08.8495.04

==EI mA

080.0101

08.8==OI mA

______________________________________________________________________________________


(a) CMO RR

υυ ⋅−

=1

21

⎟⎟⎠

⎞⎜⎜⎝

⎛+⋅⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=1

2

43

42 1

RR

RRR

CMO υυ

CMOOO RR

RR

RR

RR

υυυυ ⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛+⋅

+=+=

1

2

1

2

3

4

3

4

21 11

3

4

1

2

3

4

3

4

3

4

1

2

1

2

3

4

11

11

RR

RR

RR

RR

RR

RR

RR

RR

ACM

OCM

+

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

==υυ

(b) 3.010

69

6.94.861

4.104.62

6.94.86

=−

=+

−=CMA

(c) 03149.0

8.198.801

2.202.79

8.198.80

=+

−=CMA

or

0325.0

2.202.791

8.198.80

2.202.79

−=+

−=CMA

0325.0max

=⇒ CMA ______________________________________________________________________________________ 9.66


01

1 2 2

2

1 2 2

1 21

1 2 1 2

1 22

1 2 1 2

(1)

(2)

(3)

(4)

AI A A B

v

I B B A B

v

A I

B I

v vv v v vR R R Rv v v v vR R R R

R Rv v vR R R R

R Rv v vR R R R

−

+

−− −= +

+− −

= ++

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

( )

( )

( )

1 2 1 1 2 2

22 1

1

01

1 2 1 2 2 2

2

1 2 1 2 2

01 2

1 2 1 2 2 2 1 1 2

Now

So that

1 1 1 1

1 1 1 2

Then

1 1 1 1 1

A I B I

A B I I

I BA

V V

I AB

V V

I BB

V V

v v R v R v R v R vRv v v vR

vv vvR R R R R R R R

v vvR R R R R R R

vv v RvR R R R R R R R R R R R

− += ⇒ + = +

= + −

⎛ ⎞= + + − −⎜ ⎟+ +⎝ ⎠

⎛ ⎞= + + −⎜ ⎟+ +⎝ ⎠

⎛ ⎞ ⎛ ⎞= + + − − + +⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠⎝ ⎠

( )

( )

2 12

2 22 1

1 2 1 2 2 1

1 (1)

1 1 1 1 (2)

I IV

IB B I I

V V

v vR

v Rv v v vR R R R R R R R

⎛ ⎞+ −⎜ ⎟

⎝ ⎠⎛ ⎞ ⎡ ⎤

= + + − + −⎜ ⎟ ⎢ ⎥+ + ⎣ ⎦⎝ ⎠ Subtract (2) from (1)

( ) ( ) ( )

( )

( )

02 21 2 2 1 2 1

1 2 1 1 2 2 2 1

0 2 22 1

2 1 1 2 2 1 2 1

2 2 2 1 20 2 1

1 1 2 1 2

20

1 1 1 1 1

1 1 1 1 1

1

2

I I I I I IV V

I IV V

I IV V

vR Rv v v v v vR R R R R R R R R R

v R Rv vR R R R R R R R R R

R R R R Rv v vR R R R R R R

RvR

⎛ ⎞⎛ ⎞− = + + − − + ⋅ −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

⎧ ⎫⎛ ⎞⎛ ⎞⎪ ⎪= − + + + + ⋅⎨ ⎬⎜ ⎟⎜ ⎟ + +⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭⎧ ⎫⎛ ⎞

= − + + + +⎨ ⎬⎜ ⎟ + +⎝ ⎠⎩ ⎭

= ( )22 1

1

1 I IV

R v vR

⎛ ⎞+ −⎜ ⎟

⎝ ⎠ ______________________________________________________________________________________ 9.67

(a) ( ) ( )

titt

Ri II ω

ωωυυsin16

10sin08.02.1sin08.02.1

11

211 −=⇒

+−−=

−= ( μ A)

( ) ( )( ) tttO ωωωυ sin72.02.140sin016.0sin08.02.11 −=−−= (V) ( ) ( )( ) tttO ωωωυ sin72.02.140sin016.0sin08.02.12 +=−−+= (V)

( ) ( )( ) ttRR

OOO ωωυυυ sin32.4sin72.0240

12012

3

4 =⎟⎠⎞

⎜⎝⎛=−= (V)


)

(b) ( ) (10

sin05.060.0sin05.065.0 ttii

ωω −−−+−=

ti ωsin1051 +−= ( μ A) ( )( )ttO ωωυ sin010.0005.040sin05.065.01 +−++−= tωsin45.085.0 +−= (V) ( )( ) tttO ωωωυ sin45.040.0sin01.0005.040sin05.060.02 −−=+−−−−= (V)

( ) ( )[ ] tttO ωωωυ sin7.235.1sin45.085.0sin45.040.040

120−=+−−−−⎟

⎠⎞

⎜⎝⎛= (V)

______________________________________________________________________________________ 9.68

(a)

401 2.1667si12OB Iv v n tω⎛ ⎞= + =⎜ ⎟

⎝ ⎠

(b) 30 1.25sin12OC Iv v tω= − = −

(c)

( )2.1667sin 1.25sin3.417sin

O OB OC

O

v v v t tv t

ω ωω

= − = − −=

(d)

3.417 6.830.5

O

I

vv

= =

______________________________________________________________________________________ 9.69

(a) R

i IIO

21 υυ −=

(b) ( )Ω=⇒

−−= 100

525.025.0 RR

(c) ( )( ) 25.51525.011 =+=+= LOIO Riυυ V 25.022 −== IO υυ V

(d) 15.0

75.125.121 −=−

=−

=R

i IIO

υυmA

( )( ) 75.13125.111 −=−=+= LOIO Riυυ V 75.122 == IO υυ V

______________________________________________________________________________________ 9.70

( ) ( )

4 2

2 1 3 1

1

11

1 1 11

21

2 115200 1 0.06sin50

230For 0.5 1.0833 212.3 K

2308 V 32.33 7.11 K 7.11 K, (potentiometer) 205.2 K

Od

I I

O

O

O f

v R RAv v R R

v tR

v RR

v R R RR

ω

⎛ ⎞= = +⎜ ⎟− ⎝ ⎠

⎛ ⎞= +⎜ ⎟

⎝ ⎠

= = ⇒ =

= = ⇒ = ⇒ = =

______________________________________________________________________________________


For 10=Oυ V, 05.020010200 12 ==−⇒= IIdA υυ V

( ) ( ) 1105005.0

161 =⇒×

=−

fixedRfixedR kΩ

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

1

2

3

4 21

RR

RR

Ad

( ) 5.391

215.2200 2

2 =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+= R

RkΩ

For 5=dA

( ) ( ) ( ) ( ) ( ) 1varvar795.39215.25 11111

+=+==⇒⎥⎦

⎤⎢⎣

⎡+= RfixedRRR

R

kΩ ( ) 78var1 =R______________________________________________________________________________________ 9.72

( ) 13

41 OOO R

Rυυυ ⋅

′−=

( ) 23

4

43

42 1 OOO R

RRR

Rυυυ ⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛′

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

( ) ( 21 OOOOO )υυυυυ += and CMOO υυυ ≡= 21 Then

⎟⎟⎠

⎞⎜⎜⎝

⎛′

−⎟⎟⎠

⎞⎜⎜⎝

⎛′

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

==3

4

3

4

43

4 1RR

RR

RRR

ACM

OCM υ

υ

k , k602 34 == RR Ω 303 =R Ω , 303 =′R kΩ %5± For kΩ k 303 =′R %5− 5.28= Ω

03509.05.28

605.28

6013060

60−=⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛

+=CMA

For kΩ k 303 =′R 5.31%5 =+ Ω

03175.05.31

605.31

6013060

60+=⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛

+=CMA

Then 03175.003509.0 +≤≤− CMA______________________________________________________________________________________ 9.73

(a) s ( )( ) 46321 1041002.01020 −− ×=××=CR

( ) ( ) ttdtO ωω

ωυ sin104

25.0cos25.01041

44⋅

×−

=×−

=−− ∫

For ( )( ) 3981210425.0 4 =⇒=×⇒= − ffO πυ Hz Phase °= 90


(b) (i) ( ) 3.661042

25.05.14

=⇒×

==−

ffO π

υ Hz

(ii) ( ) 6631042

25.015.04

=⇒×

==−

ffO π

υ Hz

______________________________________________________________________________________ 9.74

(a) ( ) ( )2.1

021021

25.011 tCR

tdtCR

t

IO ′−=′′−

= ∫υυ

( )( ) 602.125.05 2121

=⇒−

=− CRCR

ms

(b) (i) ( ) 306.010.050 =′′⇒′′⋅+−= tt s, 2.4=t s

(ii) ( ) 606.010.055 =′′⇒′′⋅+−= tt s, 2.7=t s

______________________________________________________________________________________ 9.75

(a) 1

2

RZ

A−

=υ , where 22

2

22

22

222 11

11

CRjR

CjR

CjR

CjRZ

ωω

ωω +

=+

⎟⎟⎠

⎞⎜⎜⎝

⎛

==

221

2

11

CRjRR

Aωυ +

⋅−

=

(b) At 0=ω , ( )1

20RR

A−

=υ

(c) ( )2

221

2

1

1

CRRR

Aω

υ+

⋅=

Set ( )2222

222 2

1121CR

fCR

CRπ

ωω =⇒=⇒=+

______________________________________________________________________________________ 9.76

(a) kΩ 201 =R

30015 21

2 =⇒= RRR

k Ω

fCR

πω 21

22

==

( )( ) 106103001052

12

1233

22 =⇒

××== C

fRC

ππpF


(b) kΩ 151 =R

37525 21

2 =⇒= RRR

kΩ

( )( ) 3.281037510152

12332 =⇒

××= CC

πpF

______________________________________________________________________________________ 9.77

(a) 1

2

ZR

A−

=υ , where 1

11

111

11Cj

CRjCj

RZωω

ω+

=+=

11

11

1

2

11

12

11 CRjCRj

RR

CRjCRj

Aω

ωωω

υ +⋅

−=

+−

=

(b) As ∞⇒ω , 1

2

RR

A−

=υ

(c) ( )2

11

11

1

2

1 CR

CRRR

Aω

ωυ

+⋅=

Set ( ) 11112

11

11

211

21

1 CRf

CRCR

CRπ

ωω

ω=⇒=⇒=

+

______________________________________________________________________________________ 9.78

(a) Set kΩ 3502 =R

33.2315 11

2 =⇒= RRR

kΩ

( )( ) 34110201033.232

12

112 1331

111

=⇒××

==⇒= CfR

CCR

fππ

π pF

(b) Set k 201 =R Ω

50025 21

2 =⇒= RRR

kΩ

( )( ) 227103510202

11331 =⇒

××= CC

πpF

______________________________________________________________________________________ 9.79 Assuming the Zener diode is in breakdown,

( )

( )

2

1

2 22

2

1 6.8 6.8 1

0 6.80 6.8 1

10 10 6.8 6.8 6.2 !!!5.6

O z O

O

zz z

s

Rv V v VR

vi i mAR

Vi i iR

= − ⋅ = − ⇒ = −

− −−= = ⇒ =

− −= − = − ⇒ = − mA

Circuit is not in breakdown. Now


( )( )

2 21

2 2

10 0 10 1.52 5.6 11.52 1 1.52

0

s

O O

z

i i mR Rv i R v

i

−= = ⇒ =

+ += − = − ⇒ = −

=

A

V

______________________________________________________________________________________ 9.80

( ) ( )( ) 1014 41

ln 0.026 ln 0.026ln1010 10

For 20 , 0.497 For 2 , 0.617

I IO T O

s

I O

I O

v vv V vI R

v mV v Vv V v V

−−

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − = − ⇒ = −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎣ ⎦

= == =

Iv

______________________________________________________________________________________


( ) ( )

( )( )

0 01 02 01 0

101 1

202 2

1 201 02

2 1

2 12 1

2 1

2 101 02

2 1

20

1

333 16.6520

ln

ln

ln ln

,

So ln

Then

16.65 0.026 ln

CBE T

S

CBE T

S

C CT T

C C

C C

T

v v v v v

iv v VI

iv v VI

i iv v V V

i iv vi iR R

v Rv v VR v

v Rvv

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

⎛ ⎞= − = − ⎜ ⎟

⎝ ⎠⎛ ⎞

= − = − ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞− = − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= =

⎛ ⎞− = ⋅⎜ ⎟

⎝ ⎠

= ⋅

2

( ) ( ) ( ) ( )( )

( )

1

2

2 10

1 2

10

10

2 10 10

1 2

0.4329ln

ln log log log 10

2.3026log

Then 1.0 log

e e

R

v Rvv R

x x x

x

v Rvv R

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠= = ⋅⎡ ⎤ ⎡⎣ ⎦ ⎣=

⎛ ⎞≅ ⋅⎜ ⎟

⎝ ⎠

⎤⎦

______________________________________________________________________________________ 9.82

( ) ( )( )( )

/ /14 4

/ 0.02610

5

10 10

10

For 0.30 , 1.03 10 For 0.60 , 1.05

I T I

I

v V v VO s

vO

I o

I o

v I R e e

v e

v V v Vv V v V

−

−

−

= − = −

=

= = ×= =

T

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.83 From Figure 9.40

⎥⎦

⎤⎢⎣

⎡⋅+⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛++⋅−⋅−= 314

22

1

1 IB

PI

A

P

N

FI

FI

FO R

RRR

RR

RR

RR

υυυυυ

3142 3210 IIII υυυυ ++−−=

Then 101

=RRF , 1

2

=RRF , 21 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

A

P

N

F

RR

RR

, 31 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

B

P

N

F

RR

RR

Set k , kΩ , 500=FR Ω 501 =R 5002 =R kΩ Now 45.455005021 === RRRN kΩ

Then 21245.45

5001 =⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +

A

P

A

P

RR

RR

, Also 31245.45

5001 =⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +

B

P

B

P

RR

RR

Let k , then 500=AR Ω 3.33332

== AB RR kΩ

Then kΩ33.83=PR CBA RRR=

We find 2003.333500 ==BA RR kΩ

So 8.14233.83200 =⇒= CC RR kΩ ______________________________________________________________________________________ 9.84

52

41

3211 IF

IR

IC

PI

B

PI

A

P

N

FO R

RRR

RR

RR

RR

RR

υυυυυυ ⋅−⋅−⎥⎦

⎤⎢⎣

⎡⋅+⋅+⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

54321 6425.13 IIIII υυυυυ −−++=

We have 41

=RRF , 6

2

=RRF ; Set 250=FR kΩ , 5.621 =R kΩ , 67.412 =R k Ω

Now 2567.415.6221 === RRRN kΩ

Also 1125

25011 =⎟⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

N

F

RR

Now ( )

311

=A

P

RR

, ( )

5.111

=B

P

RR

, ( )

211

=C

P

RR

21

=⇒B

A

RR

, 32

=C

A

RR

Set k , k , 250=BR Ω 125=AR Ω 5.187=CR kΩ This yields k , We have 09.34=PR Ω DCBAp RRRRR =

We find 69.575.187250125 ==CBA RRR kΩ

Then 3.8309.3469.57 =⇒= DD RR kΩ ______________________________________________________________________________________ 9.85

143.16.5

1211

2

1

2 =⇒=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

VV

Z

O

F

ZOF R

VVI

−= ; Set ( ) 2.1min == ZF II mA

Then 33.52.1

6.512=

−=FR kΩ

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Set mA 15.01 =DI V 3.67.06.5 =+=+=′ γVVV ZZ

Let mA, 2.04 =I 5.312.03.6

4 ==⇒ R kΩ

Then mA 35.015.02.03 =+=I

So 6.1035.0

3.610

33 =

−=

′−=

IVV

R ZS kΩ

______________________________________________________________________________________ 9.86

2.32

6.5121 =

−=

−=

Z

ZO

IVV

R kΩ

143.16.5

1213

2

3

2 =⇒=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

RR

RR

VV

Z

O

Let mA, 2=RI 62

1232 ===+⇒

R

O

IV

RR kΩ

Then , k6143.1 33 =+ RR 8.23 =⇒ R Ω and 2.32 =R kΩ

Let mA, 44 =RI 75.04

1215

44 =

−=

−=

R

OIN

IVV

R kΩ

______________________________________________________________________________________ 9.87 Let k 20321 === RRR Ω Let ( )δ+= 120TR kΩ

Now ( ) 51021

13

31 ==⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+

== +VRR

RAO υυ V

( )( ) ( ) ( )

δδ

δδυυ

++

=⎥⎦

⎤⎢⎣

⎡++

+=⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+

== +

211010

20120120

22 V

RRR

T

TBO

So ( ) ( ) ( ) δδδ

δδδδυυυ 5.2

25

211025

21105 −=−≅

++−+

=++

−=−= BAOA

We have ⎟⎠⎞

⎜⎝⎛ −

=300

300Tyδ ; At , 350=T 21=TR kΩ , ( ) 05.012021 =⇒+=⇒ δδ

Then 30.0300

30035005.0 =⇒⎟⎠⎞

⎜⎝⎛ −

= yy

For 05.0=δ , ( )( ) 125.005.05.2 ==OAυ V

For the instrumentation amplifier, ( )125.02

151

2

3

4⎟⎟⎠

⎞⎜⎜⎝

⎛+==

RR

RR

Oυ

For example, set 43

4 =RR

and 5.41

2 =RR

______________________________________________________________________________________


(a) ++ ⋅⎟⎠⎞

⎜⎝⎛ Δ−

=⋅⎟⎠⎞

⎜⎝⎛

Δ++Δ−Δ−

= VR

RRVRRRR

RRA 2

υ

++ ⋅⎟⎠⎞

⎜⎝⎛ Δ+

=⋅⎟⎠⎞

⎜⎝⎛

Δ−+Δ+Δ+

= VR

RRVRRRR

RRB 2

υ

⎟⎠⎞

⎜⎝⎛

×Δ−=⋅

Δ−=⋅⎥⎦

⎤⎢⎣⎡ Δ+

−Δ−

=−= ++31 1020

922

RVRRV

RRR

RRR

BAO υυυ

or ( )( )RO Δ×−= −41 105.4υ

(b) For an instrumentation amplifier,

11

2

3

4 21 OO R

RRR

υυ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+=

For , Ω=Δ 200R 5−=Oυ V

( )( )200105.42

15 4

1

2

3

4 −×−⎟⎟⎠

⎞⎜⎜⎝

⎛+=−

RR

RR

or ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

1

2

3

4 2155.55

RR

RR

For example, set 63

4 =RR

and 13.41

2 =RR

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Micro4ProbSol9

Documents

r4 r3

r1 r2

r1 r3

r2 r1

r1 r1

r1 r1

v2 v1

instrumentation